Chapter 22 The Electric Field II: Continuous Charge Distributions

Transcription

1 Chapte The lectic Field II: Continuous Chage Distibutions 1 [M] A unifom line chage that has a linea chage density l equal to.5 nc/m is on the x axis between x and x 5. m. (a) What is its total chage? Find the electic field on the x axis at (b) x 6. m, (c) x 9. m, and (d) x 5 m. (e) stimate the electic field at x 5 m, using the appoximation that the chage is a point chage on the x axis at x.5 m, and compae you esult with the esult calculated in Pat (d). (To do this you will need to assume that the values given in this poblem statement ae valid to moe than two significant figues.) Is you appoximate esult geate o smalle than the exact esult? xplain you answe. Pictue the Poblem We can use the definition of λ to find the total chage of the line of chage and the expession fo the electic field on the axis of a finite line of chage to evaluate x at the given locations along the x axis. In Pat (d) we can apply Coulomb s law fo the electic field due to a point chage to appoximate the electic field at x 5 m. (a) Use the definition of linea chage density to expess in tems of λ: λl 18nC (.5nC/m)( 5.m) 17.5nC xpess the electic field on the axis of a finite line chage: x ( x ) x k ( x L) (b) ubstitute numeical values and evaluate x at x 6. m: x 9 ( ) ( N m /C )( 17.5nC) 6.m ( 6.m)( 6.m 5.m) 6 N/C (c) ubstitute numeical values and evaluate x at x 9. m: x 9 ( ) ( N m /C )( 17.5nC) 9.m ( 9.m)( 9.m 5.m) 4.4 N/C (d) ubstitute numeical values and evaluate x at x 5 m: x 9 ( ) ( N m /C )( 17.5nC) 5m ( 5m)( 5m 5.m).568 mn/c.6mn/c 89

2 9 Chapte (e) Use Coulomb s law fo the electic field due to a point chage to obtain: k x ( x) x

3 The lectic Field II: Continuous Chage Distibutions 91 ubstitute numeical values and evaluate x (5 m): x 9 ( ) ( N m /C )( 17.5nC) 5m ( 5m.5 m).56774mn/c.6 mn/c This esult is about.1% less than the exact value obtained in (d). This suggests that the line of chage is too long fo its field at a distance of 5 m to be modeled exactly as that due to a point chage. 17 [M] A ing that has adius a lies in the z plane with its cente at the oigin. The ing is unifomly chaged and has a total chage. Find z on the z axis at (a) z.a, (b) z.5a, (c) z.7a, (d) z a, and (e) z a. (f) Use you esults to plot z vesus z fo both positive and negative values of z. (Assume that these distances ae exact.) z Pictue the Poblem We can use z πkq 1 to find the electic z + a field at the given distances fom the cente of the chaged ing. (a) valuate z (.a): (.a) (b) valuate z (.5a): (.5a) (c) valuate z (.7a): (.7a) z z z k(.a) (.a) + a [ ] k.189 a k(.5a) (.5a) + a [ ] k.58 a k(.7a) (.7a) + a [ ] k.85 a (d) valuate z (a): ka z ( a) k.54 [ a + a ] a

4 9 Chapte (e) valuate z (a): ka z ( a) k.179 [( a) + a ] a

5 The lectic Field II: Continuous Chage Distibutions 9 (f) The field along the x axis is plotted below. The z coodinates ae in units of z/a and is in units of k/a..4. x z/a 18 A non-conducting disk of adius a lies in the z plane with its cente at the oigin. The disk is unifomly chaged and has a total chage. Find z on the z axis at (a) z.a, (b) z.5a, (c) z.7a, (d) z a, and (e) z a. (f) Use you esults to plot z vesus z fo both positive and negative values of z. (Assume that these distances ae exact.) z Pictue the Poblem We can use z π kq 1, whee a is the adius z + a of the disk, to find the electic field on the axis of a chaged disk. The electic field on the axis of a chaged disk of adius a is given by: z πk 1 1 z z z + a z + a (a) valuate z (.a): z (.a) 1.4.a (.a) + a

6 94 Chapte (b) valuate z (.5a): z (.5a) a (.5a) + a (c) valuate z (.7a): z (.7a) 1.1.7a (.7a) + a (d) valuate z (a): z ( a) a a + a (e) valuate z (a): z ( a) 1.58 a ( a) + a The field along the x axis is plotted below. The x coodinates ae in units of z/a and is in units of x z/a

7 The lectic Field II: Continuous Chage Distibutions 95 7 A squae that has 1-cm-long edges is centeed on the x axis in a ρ egion whee thee exists a unifom electic field given by (. kn/c)iˆ. (a) What is the electic flux of this electic field though the suface of a squae if the nomal to the suface is in the +x diection? (b) What is the electic flux though the same squae suface if the nomal to the suface makes a 6º angle with the y axis and an angle of 9 with the z axis? Pictue the Poblem The definition of electic flux isφ ρ nˆda. We can apply this definition to find the electic flux though the squae in its two oientations. (a) Apply the definition of φ to find the flux of the field when the squae is paallel to the yz plane: φ (.kn/c) (.kn/c) (.kn/c)(.1m). N m iˆ ida ˆ da /C φ (. kn/c) i : (. kn/c) (b) Poceed as in (a) with ˆ n ˆ cos (. kn/c)(.1 m) 17 N m /C cos da cos da cos ρ 9 [M] An electic field is given by sign( x) ( N/C)iˆ, whee sign(x) equals 1 if x <, if x, and +1 if x >. A cylinde of length cm and adius 4. cm has its cente at the oigin and its axis along the x axis such that one end is at x +1 cm and the othe is at x 1 cm. (a) What is the electic flux though each end? (b) What is the electic flux though the cuved suface of the cylinde? (c) What is the electic flux though the entie closed suface? (d) What is the net chage the cylinde? Pictue the Poblem The field at both cicula faces of the cylinde is paallel to the outwad vecto nomal to the suface, so the flux is just A. Thee is no flux though the cuved suface because the nomal to that suface is pependicula to ρ. The net flux though the closed suface is elated to the net chage by Gauss s law.

8 96 Chapte (a) Use Gauss s law to calculate the flux though the ight cicula suface: Apply Gauss s law to the left cicula suface: φ φ ight left ρ ρ ight nˆ ight ( N/C) iˆ iˆ ( π )(.4 m) 1.5 N m left nˆ left A A /C ( N/C) iˆ ( iˆ )( π )(.4 m) 1.5 N m /C (b) Because the field lines ae paallel to the cuved suface of the cylinde: φ cuved (c) xpess and evaluate the net flux though the entie cylindical suface: φ net φ + φ + φ ight left 1.5 N m /C N m /C +. N m /C cuved (d) Apply Gauss s law to obtain: φ net 4πk φnet 4πk ubstitute numeical values and evaluate : 4 π 9 ( N m /C ).7 1. N m 11 C /C

9 The lectic Field II: Continuous Chage Distibutions 97 1 A point chage (q +. μc) is at the cente of an imaginay sphee that has a adius equal to.5 m. (a) Find the suface aea of the sphee. (b) Find the magnitude of the electic field at all points on the suface of the sphee. (c) What is the flux of the electic field though the suface of the sphee? (d) Would you answe to Pat (c) change if the point chage wee moved so that it was the sphee but not at its cente? (e) What is the flux of the electic field though the suface of an imaginay cube that has 1.-m-long edges and encloses the sphee? Pictue the Poblem We can apply Gauss s law to find the flux of the electic field though the suface of the sphee. (a) Use the fomula fo the suface aea of a sphee to obtain: A 4π 4π.14m (.5 m).14 m (b) Apply Coulomb s law to find : 4π 1 q 4π N/C 1.μC 1 ( C /N m )(.5 m) N/C (c) Apply Gauss s law to obtain: nda ˆ da 4 ( N/C)(.14 m ) φ ρ N m /C (d) No. The flux though the suface is independent of whee the chage is located the sphee. (e) Because the cube encloses the sphee, the flux though the suface of the sphee will also be the flux though the cube: φ cube N m /C 4 Because the fomulas fo Newton s law of gavity and fo Coulomb s law have the same invese-squae dependence on distance, a fomula analogous to the fomula fo Gauss s law can be found fo gavity. The gavitational field g ρ at a location is the foce pe unit mass on a test mass m placed at that location. Then, fo a point mass m at the oigin, the gavitational field g ρ at some position

10 98 Chapte ( ρ ) is g ρ ( Gm )ˆ. Compute the flux of the gavitational field though a spheical suface of adius R centeed at the oigin, and veify that the gavitational analog of Gauss s law is φ net 4πGm. Pictue the Poblem We ll define the flux of the gavitational field in a manne that is analogous to the definition of the flux of the electic field and then substitute fo the gavitational field and evaluate the integal ove the closed spheical suface. Define the gavitational flux as: φ g ρ nˆda g ubstitute fo g ρ and evaluate the integal to obtain: φ net Gm Gm Gm 4πGm ˆ nˆ da da ( 4π ) 4 Conside the solid conducting sphee and the concentic conducting spheical shell in Figue -41. The spheical shell has a chage 7. The solid sphee has a chage +. (a) How much chage is on the oute suface and how much chage is on the inne suface of the spheical shell? (b) uppose a metal wie is now connected between the solid sphee and the shell. Afte electostatic equilibium is e-established, how much chage is on the solid sphee and on each suface of the spheical shell? Does the electic field at the suface of the solid sphee change when the wie is connected? If so, in what way? (c) uppose we etun to the conditions in Pat (a), with + on the solid sphee and 7 on the spheical shell. We next connect the solid sphee to gound with a metal wie, and then disconnect it. Then how much total chage is on the solid sphee and on each suface of the spheical shell? Detemine the Concept The chages on a conducting sphee, in esponse to the epulsive Coulomb foces each expeiences, will sepaate until electostatic equilibium conditions exit. The use of a wie to connect the two sphees o to gound the oute sphee will cause additional edistibution of chage. (a) Because the oute sphee is conducting, the field in the thin shell must vanish. Theefoe,, unifomly distibuted, esides on the inne suface, and 5, unifomly distibuted, esides on the oute suface.

11 The lectic Field II: Continuous Chage Distibutions 99

12 1 Chapte (b) Now thee is no chage on the inne suface and 5 on the oute suface of the spheical shell. The electic field just outside the suface of the inne sphee changes fom a finite value to zeo. (c) In this case, the 5 is dained off, leaving no chage on the oute suface and on the inne suface. The total chage on the oute sphee is then. 41 A non-conducting solid sphee of adius 1. cm has a unifom volume chage density. The magnitude of the electic field at. cm fom the sphee s cente is N/C. (a) What is the sphee s volume chage density? (b) Find the magnitude of the electic field at a distance of 5. cm fom the sphee s cente. Pictue the Poblem (a) We can use the definition of volume chage density, in conjunction with quation -18a, to find the sphee s volume chage density. (b) We can use quation -18b, in conjunction with ou esult fom Pat (a), to find the electic field at a distance of 5. cm fom the solid sphee s cente. (a) The solid sphee s volume chage density is the atio of its chage to its volume: (1) V ρ 4 πr Fo R, quation -18a gives the electic field at a distance fom the cente of the sphee: 1 () 4π olving fo yields: 4π ubstitute fo in equation (1) and simplify to obtain: 4π ρ 4 πr R ubstitute numeical values and evaluate ρ: 1 ( 1 C /N m )( N/C)(. cm) ρ. μc/m ( 1. cm) 1.997μC/m (b) Fo R, the electic field at a distance fom the cente of the sphee is given by: 1 () 4π R

13 The lectic Field II: Continuous Chage Distibutions 11 xpess fo R: 4 ρv π ρ sphee adius is whose ubstituting fo in equation () and simplifying yields: 1 4π 4 π R 4 ρ ρ R ubstitute numeical values and evaluate (5. cm): 4 ( ) ( 1.997μC/m )( 5. cm). cm 1 ( C /N m )( 1. cm) 5 47 N/C 4 [M] A sphee of adius R has volume chage density ρ B/ fo < R, whee B is a constant and ρ fo > R. (a) Find the total chage on the sphee. (b) Find the expessions fo the electic field and outside the chage distibution (c) ketch the magnitude of the electic field as a function of the distance fom the sphee s cente. Pictue the Poblem We can find the total chage on the sphee by expessing the chage dq in a spheical shell and integating this expession between and R. By symmety, the electic fields must be adial. To find the chaged sphee we choose a spheical Gaussian suface of adius < R. To find outside the chaged sphee we choose a spheical Gaussian suface of adius > R. On each of these sufaces, is constant. Gauss s law then elates to the total chage the suface. (a) xpess the chage dq in a shell of thickness d and volume 4π d: Integate this expession fom to R to find the total chage on the sphee: dq 4π ρd 4π 4πBd R 4πB d πbr B [ πb ] d R (b) Apply Gauss s law to a spheical suface of adius > R that is concentic with the nonconducting sphee to obtain: 1 da o 4π

14 1 Chapte olving fo yields: ( > R) 4π 1 k kπbr BR Apply Gauss s law to a spheical suface of adius < R that is concentic with the nonconducting sphee to obtain: 1 da 4π olving fo yields: ( < R) 4π B πb 4π

15 The lectic Field II: Continuous Chage Distibutions 1 (c) The following gaph of vesus /R, with in units of B/( ), was plotted using a speadsheet pogam /R Remaks: Note that ou esults fo (a) and (b) agee at R. 46 Fo you senio poject you ae in chage of designing a Geige tube fo detecting adiation in the nuclea physics laboatoy. This instument will consist of a long metal cylindical tube that has a long staight metal wie unning down its cental axis. The diamete of the wie is to be.5 mm and the diamete of the tube will be 4. cm. The tube is to be filled with a dilute gas in which electical dischage (beakdown) occus when the electic field eaches N/C. Detemine the maximum linea chage density on the wie if beakdown of the gas is not to happen. Assume that the tube and the wie ae infinitely long. Pictue the Poblem The electic field of a line chage of infinite length is given 1 λ by, whee is the distance fom the cente of the line of chage and π λ is the linea chage density of the wie. The electic field of a line chage of infinite length is given by: 1 λ π Because vaies invesely with, its maximum value occus at the suface of the wie whee R, the adius of the wie: max 1 λ R π olving fo λ yields: λ π R max ubstitute numeical values and evaluate λ:

16 14 Chapte λ π C N m N C 6 (.5 mm) nc/m 48 how that the electic field due to an infinitely long, unifomly chaged thin cylindical shell of adius a having a suface chage density is given by the following expessions: fo R < a and R a R ( ) fo R > a. Pictue the Poblem Fom symmety, the field in the tangential diection must vanish. We can constuct a Gaussian suface in the shape of a cylinde of adius and length L and apply Gauss s law to find the electic field as a function of the distance fom the centeline of the infinitely long, unifomly chaged cylindical shell. Apply Gauss s law to the cylindical suface of adius and length L that is concentic with the infinitely long, unifomly chaged cylindical shell: 1 n da o πl R whee we ve neglected the end aeas because no thee is no flux though them. olve fo R : R πl k L Fo < R, and: ( < R) Fo > R, λl and: ( > R) R R kλl kλ k L R ( πr ) 49 A thin cylindical shell of length m and adius 6. cm has a unifom suface chage density of 9. nc/m. (a) What is the total chage on the shell? Find the electic field at the following adial distances fom the long axis of the cylinde. (b). cm, (c) 5.9 cm, (d) 6.1 cm, and (e) 1. cm. (Use the esults of Poblem 48.) Pictue the Poblem We can use the definition of suface chage density to find the total chage on the shell. Fom symmety, the electic field in the tangential

17 The lectic Field II: Continuous Chage Distibutions 15 diection must vanish. We can constuct a Gaussian suface in the shape of a cylinde of adius and length L and apply Gauss s law to find the electic field as a function of the distance fom the centeline of the unifomly chaged cylindical shell. (a) Using its definition, elate the suface chage density to the total chage on the shell: ubstitute numeical values and evaluate : A πrl π (.6 m)( m)( 9. nc/m ) 679nC (b) Fom Poblem 48 we have, fo. cm: (c) Fom Poblem 48 we have, fo 5.9 cm: (.cm) ( 5.9cm) (d) Fom Poblem 48 we have, fo 6.1 cm: and () R ( ) ( 6 9. nc/m )(.6 m).1cm 1 ( C /N m )(.61 m) 1. kn/c

18 16 Chapte (e) Fom Poblem 48 we have, fo 1. cm: ( ) ( 1 9. nc/m )(.6 m).cm 1 ( C /N m )(.1 m) 61 N/C 5 An infinitely long non-conducting solid cylinde of adius a has a unifom volume chage density of ρ. how that the electic field is given by the following expessions: R ρ R ( ) fo R < a and R ρ a ( R) fo R > a, whee R is the distance fom the long axis of the cylinde. Pictue the Poblem Fom symmety, the field tangent to the suface of the cylinde must vanish. We can constuct a Gaussian suface in the shape of a cylinde of adius and length L and apply Gauss s law to find the electic field as a function of the distance fom the centeline of the infinitely long nonconducting cylinde. Apply Gauss s law to a cylindical suface of adius and length L that is concentic with the infinitely long nonconducting cylinde: 1 n da o πl R whee we ve neglected the end aeas because thee is no flux though them. olving fo R yields: R πl k L xpess fo < R: ρ ( ) V ρ( π L) ubstitute to obtain: ( πρ L ) k ρ R ( < R) L o, because λ ρπr, λ π R ( < R) R xpess fo > R: ρ ( ) V ρ( πr L)

19 The lectic Field II: Continuous Chage Distibutions 17 ubstitute fo to obtain: ( πρ LR ) k ρr R ( > R) L o, because λ ρπr ( > R) λ π R 5 Figue -4 shows a potion of an infinitely long, concentic cable in coss section. The inne conducto has a chage of 6. nc/m and the oute conducto has no net chage. (a) Find the electic field fo all values of R, whee R is the pependicula distance fom the common axis of the cylindical system. (b) What ae the suface chage densities on the and the outside sufaces of the oute conducto? Pictue the Poblem The electic field is diected adially outwad. We can constuct a Gaussian suface in the shape of a cylinde of adius and length L and apply Gauss s law to find the electic field as a function of the distance fom the centeline of the infinitely long, unifomly chaged cylindical shell. (a) Apply Gauss s law to a cylindical suface of adius and length L that is concentic with the inne conducto: 1 n da πl R whee we ve neglected the end aeas because thee is no flux though them. olving fo R yields: R k (1) L Fo < 1.5 cm, and: R ( <1.5cm) Letting R 1.5 cm, expess fo 1.5 cm < < 4.5 cm: λl πrl ubstitute in equation (1) to obtain: k R ( 1.5cm < < 4.5cm) L kλ ubstitute numeical values and evaluate n (1.5 cm < < 4.5 cm): R ( λl) ( ) ( ) ( 6. nc/m ) ( 18N m/c 1.5cm 4.5cm N m /C ) 9 < <

20 18 Chapte xpess fo 4.5 cm < < 6.5 cm: Letting epesent the chage density on the oute suface, expess fo > 6.5 cm: ubstitute in equation (1) to obtain: and R ( 4.5cm < < 6.5cm) A π RL whee R 6.5 cm. ( π R L) k R R ( > R ) L In (b) we show that 1. nc/m. ubstitute numeical values to obtain: R ( ) ( 1. nc/m )( 6.5cm).5cm > N m/c 1 ( C / N m )

21 The lectic Field II: Continuous Chage Distibutions 19 (b) The suface chage densities on the and the outside sufaces of the oute conducto ae given by: λ πr and λ outside πr outside ubstitute numeical values and 6. nc/m π evaluate and outside : (.45 m) 1. nc/m and 6. nc/m outside π (.65 m) 14.7 nc/m 1. nc/m 59 A thin metal slab has a net chage of zeo and has squae faces that have 1-cm-long sides. It is in a egion that has a unifom electic field that is pependicula to its faces. The total chage induced on one of the faces is 1. nc. What is the magnitude of the electic field? Pictue the Poblem Because the metal slab is in an extenal electic field, it will have chages of opposite signs induced on its faces. The induced chage is elated to the electic field by. / Relate the magnitude of the electic field to the chage density on the metal slab: Use its definition to expess : ubstitute fo to obtain: A L L ubstitute numeical values and 1 evaluate : (.1m) ( C /N m ) 9.4 kn/c 1.nC

22 11 Chapte 61 A conducting spheical shell that has zeo net chage has an inne adius R 1 and an oute adius R. A positive point chage q is placed at the cente of the shell. (a) Use Gauss s law and the popeties of conductos in electostatic equilibium to find the electic field in the thee egions: < R 1, R 1 < < R, and > R, whee is the distance fom the cente. (b) Daw the electic field lines in all thee egions. (c) Find the chage density on the inne suface ( R 1 ) and on the oute suface ( R ) of the shell. Pictue the Poblem We can constuct a Gaussian suface in the shape of a sphee of adius with the same cente as the shell and apply Gauss s law to find the electic field as a function of the distance fom this point. The inne and oute sufaces of the shell will have chages induced on them by the chage q at the cente of the shell. (a) Apply Gauss s law to a spheical suface of adius that is concentic with the point chage: olving fo Fo < R 1, yields: q. ubstitute in equation (1) and simplify to obtain: Because the spheical shell is a conducto, a chage q will be induced on its inne suface. Hence, fo R 1 < < R : Fo > R, q. ubstitute in equation (1) and simplify to obtain: 1 n da 4π (1) 4π q kq ( < R1 ) 4π and ( R < < R ) 1 q kq ( > R ) 4π

23 (b) The electic field lines ae shown in the diagam to the ight: The lectic Field II: Continuous Chage Distibutions 111 (c) A chage q is induced on the inne suface. Use the definition of suface chage density to obtain: inne q 4πR 1 A chage q is induced on the oute suface. Use the definition of suface chage density to obtain: oute q 4πR 6 The electic field just above the suface of ath has been measued to typically be 15 N/C pointing downwad. (a) What is the sign of the net chage on ath s suface unde typical conditions? (b)what is the total chage on ath s suface implied by this measuement? Pictue the Poblem We can constuct a spheical Gaussian suface at the suface of ath (we ll assume ath is a sphee) and apply Gauss s law to elate the electic field to its total chage. (a) Because the diection of an electic field is the diection of the foce acting on a positively chaged object, the net chage on ath s suface must be negative. (b)apply Gauss s law to a spheical suface of adius R that is concentic with ath: 1 n da 4πR n olve fo to obtain: ath ath 4 π Rn R k n ubstitute numeical values and evaluate : ath ath 6 ( m) ( 15 N/C) kc 9 N m /C

24 11 Chapte 6 [M] A positive point chage of.5 μc is at the cente of a conducting spheical shell that has a net chage of zeo, an inne adius equal to 6 cm, and an oute adius equal to 9 cm. (a) Find the chage densities on the inne and oute sufaces of the shell and the total chage on each suface. (b) Find the electic field eveywhee. (c) Repeat Pat (a) and Pat (b) with a net chage of +.5 μc placed on the shell. Pictue the Poblem Let the inne and oute adii of the unchaged spheical conducting shell be R 1 and R and q epesent the positive point chage at the cente of the shell. The positive point chage at the cente will induce a negative chage on the inne suface of the shell and, because the shell is unchaged, an equal positive chage will be induced on its oute suface. To solve Pat (b), we can constuct a Gaussian suface in the shape of a sphee of adius with the same cente as the shell and apply Gauss s law to find the electic field as a function of the distance fom this point. In Pat (c) we can use a simila stategy with the additional chage placed on the shell. (a) xpess the chage density on the inne suface: xpess the elationship between the positive point chage q and the chage induced on the inne suface q : inne q inne inne A q + q inne q q inne ubstitute fo qinne and A to obtain: inne q 4πR 1 ubstitute numeical values and.5μc π inne evaluate inne : 4 (.6m).55 μc/m xpess the chage density on the oute suface: Because the spheical shell is unchaged: q oute oute A q oute + qinne ubstitute fo q oute to obtain: oute q 4πR inne ubstitute numeical values and.5μc π oute evaluate oute : 4 (.9m).5 μc/m

25 The lectic Field II: Continuous Chage Distibutions 11 (b) Apply Gauss s law to a spheical suface of adius that is concentic with the point chage: olve fo : 1 n da 4π (1) 4π Fo < R 1 6 cm, q. ubstitute in equation (1) and evaluate ( < 6 cm) to obtain: 9 q kq ( ) ( N m /C )(.5μC) < 6 cm 4π 4 1 (. 1 N m /C)

26 114 Chapte Because the spheical shell is a conducto, a chage q will be induced on its inne suface. Hence, fo 6 cm < < 9 cm: and ( 6 cm < < 9cm) Fo > 9 cm, the net chage the Gaussian suface is q and: kq 1 4 ( > 9cm) (. 1 N m /C) (c) Because in the conducto: q inne and inne.5μc.55 μc/m as befoe. xpess the elationship between the chages on the inne and oute sufaces of the spheical shell: q oute and q oute + q inne.5μc.5μ C - q 6.μC inne oute is now given by: oute 6.μC 4 π (.9m).59 μc/m Fo < R 1 6 cm, q and ( < 6 cm) is as it was in (a): Because the spheical shell is a conducto, a chage q will be induced on its inne suface. Hence, fo 6 cm < < 9 cm: ( < 6 cm) (. 1 N m /C) and ( 6 cm < < 9cm) 4 1 Fo >.9 m, the net chage the Gaussian suface is 6. μc and: kq 9 4 ( > 9cm) ( N m /C )( 6.μ C) ( N m /C) Conside the concentic metal sphee and spheical shells that ae shown in Figue -4. The innemost is a solid sphee that has a adius R 1. A spheical shell suounds the sphee and has an inne adius R and an oute adius R. The sphee and the shell ae both suounded by a second spheical shell that

27 The lectic Field II: Continuous Chage Distibutions 115 has an inne adius R 4 and an oute adius R 5. None of these thee objects initially have a net chage. Then, a negative chage is placed on the inne sphee and a positive chage + is placed on the outemost shell. (a) Afte the chages have eached equilibium, what will be the diection of the electic field between the inne sphee and the middle shell? (b) What will be the chage on the inne suface of the middle shell? (c) What will be the chage on the oute suface of the middle shell? (d) What will be the chage on the inne suface of the outemost shell? (e) What will be the chage on the oute suface of the outemost shell? (f) Plot as a function of fo all values of. Detemine the Concept We can detemine the diection of the electic field between sphees I and II by imagining a test chage placed between the sphees and detemining the diection of the foce acting on it. We can detemine the amount and sign of the chage on each sphee by ealizing that the chage on a given suface induces a chage of the same magnitude but opposite sign on the next suface of lage adius. (a) The chage placed on sphee III has no beaing on the electic field between sphees I and II. The field in this egion will be in the diection of the foce exeted on a test chage placed between the sphees. Because the chage at the cente is negative, the field will point towad the cente. (b) The chage on sphee I ( ) will induce a chage of the same magnitude but opposite sign on sphee II: + (c) The induction of chage + on the inne suface of sphee II will leave its oute suface with a chage of the same magnitude but opposite sign: (d) The pesence of chage on the oute suface of sphee II will induce a chage of the same magnitude but opposite sign on the inne suface of sphee III: + (e) The pesence of chage + on the inne suface of sphee III will leave the oute suface of sphee III neutal:

28 116 Chapte (f) A gaph of as a function of is shown to the ight: